Question

Para cada função abaixo, determine:

F(x)=x2-4x+5
F(x)=-x2+2x-1
F(x)=x2-2x+1

1. As raízes das funções;

2. Tipo de concavidade;

3. Ponto de vértice;

4. Gráfico.

n comentem qqr coisa pra pegar pontos se n vou denunciar​

Answer 1

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$\sf f(x)=x^2-4x+5$

1)

$\sf fac_{\!\!,}a~f(x)=0\\\sf x^2-4x+5=0~a=1~b=-4~c=5\\\sf\Delta=b^2-4ac\\\sf\Delta=(-4)^2-4\cdot1\cdot5\\\sf\Delta=16-20\\\sf\Delta=-4<0\implies n\tilde ao~tem~ra\acute izes~reais$

2)

$\sf a=1>0\implies concavidade~para~cima$

3)

$\sf x_V=-\dfrac{b}{2a}\\\sf x_V=-\dfrac{-4}{2\cdot1}=\dfrac{4}{2}=2\\\sf y_V=-\dfrac{\Delta}{4a}\\\sf y_V=-\dfrac{-4}{4\cdot1}=\dfrac{4}{4}=1\\\huge\boxed{\boxed{\boxed{\boxed{\sf V(2,1)}}}}$

4)

$\sf o~gr\acute afico~est\acute a~anexo$

$\sf f(x)=-x^2+2x-1$

1)

$\sf fac_{\!\!,}a~f(x)=0\\\sf-x^2+2x-1=0~a=-1~b=2~c=-1\\\sf\Delta=b^2-4ac\\\sf\Delta=2^2-4\cdot(-1)\cdot(-1)\\\sf\Delta=4-4\\\sf\Delta=0\implies~possui~uma~\acute unica~ra\acute iz~real\\\sf x=-\dfrac{b}{2a}\\\sf x=-\dfrac{2}{\cdot(-1)}=\dfrac{2}{2}=1$

2)

$\sf a=-1<0\implies possui~concavidade~para~baixo$

3)

$\sf x_V=-\dfrac{b}{2a}\\\sf x_V=-\dfrac{2}{2\cdot(-1)}=\dfrac{2}{2}=1\\\sf y_V=-\dfrac{\Delta}{4a}\\\sf y_V=\dfrac{0}{4\cdot(-1)}=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf V(1,0)}}}}$

4)

$\sf gr\acute afico~est\acute a~anexo.$

$\sf f(x)=x^2-2x+1$

1)

$\sf fac_{\!\!,}a~f(x)=0\\\sf x^2-2x+1=0~~a=1~b=-2~c=1\\\sf\Delta=b^2-4ac\\\sf\Delta=(-2)^2-4\cdot1\cdot1\\\sf\Delta=4-4\\\sf\Delta=0\implies possui~uma~\acute unica~ra\acute iz~real\\\sf x=-\dfrac{b}{2a}\\\sf x=-\dfrac{2}{2\cdot1}=-\dfrac{-2}{2}=1$

2)

$\sf a=1>0\implies concavidade~para~cima$

3)

$\sf x_V=-\dfrac{b}{2a}\\\sf x_V=-\dfrac{-2}{2\cdot1}=\dfrac{2}{2}=1\\\sf y_V=-\dfrac{\Delta}{4a}\\\sf y_V=\dfrac{0}{4\cdot1}=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf V(1,0)}}}}$

4)

$\sf gr\acute afico~est\acute a~anexo$